\begin{answer}
We have
    $$
    \begin{aligned}
l(\phi, \mu_0, \mu_1, \Sigma) &= \sum_{i=1}^m(\log p(x^{(i)}|y^{(i)}; \mu_0, \mu_1, \Sigma) + \log p(y^{(i)};\phi))\\
&= \sum_{i=1}^m [1\{y^{(i)} = 0\} \log \frac{1}{\sqrt{2\pi}\sigma}\exp(-\frac{(x^{(i)} - \mu_0)^2}{2\sigma^2}) + 1\{y^{(i)} = 1\} \log \frac{1}{\sqrt{2\pi}\sigma}\exp(-\frac{(x^{(i)} - \mu_1)^2}{2\sigma^2}) + y^{(i)} \log \phi + (1 - y^{(i)}\log (1- \phi))]
\end{aligned}
$$
And we set derivatives to zero:

    \[
\begin{aligned}
\frac{\partial l}{\partial \phi} = y^{(i)}\frac{1}{\phi} + (1 - y^{(i)})\frac{1}{1- \phi} = 0\\
\frac{\partial l}{\partial \mu_0} = - \sum_{i=1}^m1\{y^{(i)} = 0\} \frac{x^{(i)} - \mu_0}{\sigma^2} = 0\\
\frac{\partial l}{\partial \mu_1} = - \sum_{i=1}^m1\{y^{(i)} = 1\} \frac{x^{(i)} - \mu_1}{\sigma^2} = 0\\
\frac{\partial l}{\partial \sigma^2 } = \sum_{i=1}^m -\frac{1}{2\sigma^2} + \frac{(x^{(i)} - \mu_{y^{(i)}})^2}{2\sigma^4} = 0
\end{aligned}
\]
And from these we get the estimations.
\end{answer}

